测试 numpy 数组是否是 numpy 数组列表的成员，并将其从列表中删除

Question

When testing if a numpy array c is member of a list of numpy arrays CNTS :

import numpy as np

c = np.array([[[ 75, 763]],
              [[ 57, 763]],
              [[ 57, 749]],
              [[ 75, 749]]])

CNTS = [np.array([[[  78, 1202]],
                  [[  63, 1202]],
                  [[  63, 1187]],
                  [[  78, 1187]]]),
        np.array([[[ 75, 763]],
                  [[ 57, 763]],
                  [[ 57, 749]],
                  [[ 75, 749]]]),
        np.array([[[ 72, 742]],
                  [[ 58, 742]],
                  [[ 57, 741]],
                  [[ 57, 727]],
                  [[ 58, 726]],
                  [[ 72, 726]]]),
        np.array([[[ 66, 194]],
                  [[ 51, 194]],
                  [[ 51, 179]],
                  [[ 66, 179]]])]

print(c in CNTS)

I get:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

However, the answer is rather clear: c is exactly CNTS[1] , so c in CNTS should return True!

How to correctly test if a numpy array is member of a list of numpy arrays?

The same problem happens when removing :

CNTS.remove(c)

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Application: test if an opencv contour (numpy array) is member of a list of contours, see for example Remove an opencv contour from a list of contours .

Answer 1

You are getting the error because in essentially invokes bool(c == x) on every element x of CNTS . It's the __bool__ conversion that is raising the error:

>>> c == CNTS[1]
array([[[ True,  True]],
       [[ True,  True]],
       [[ True,  True]],
       [[ True,  True]]])

>>> bool(_)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

The same applies for removal, since it tests for equality with each element.

Containment

The solution is to use np.array_equal or apply the all method to each comparison:

any(np.array_equal(c, x) for x in CNTS)

OR

any((c == x).all() for x in CNTS)

Removal

To perform the removal, you are more interested in the index of the element than its existence. The fastest way I can think of is to iterate over the indices, using the elements of CNTS as comparison keys:

index = next((i for i, x in enumerate(CNTS) if (c == x).all()), -1)

This option short circuits quite nicely, and returns -1 as the default index rather than raising a StopIteration . You can remove the argument -1 to next if you prefer the error. If you prefer, you can replace (c == x).all() with np.array_equal(c, x) .

Now you can remove as usual:

del CNTS[index]

Answer 2

This solution could work for this case:

def arrayisin(array, list_of_arrays):
    for a in list_of_arrays:
        if np.array_equal(array, a):
            return True
    return False

This function iterates over a list of arrays and tests the equality against some other array. So the usage would be:

>>> arrayisin(c, CNTS)
True

To remove the array from the list, you can get the index of the array and then use list.pop . In the function get_index , we enumerate the list of arrays, meaning we zip the indices of the list and the contents of the list. If there is a match, we return the index of the match.

def get_index(array, list_of_arrays):
    for j, a in enumerate(list_of_arrays):
        if np.array_equal(array, a):
            return j
    return None

idx = get_index(c, CNTS)  # 1
CNTS.pop(idx)

Please see the python data structures tutorial for the documentation of list.pop https://docs.python.org/3/tutorial/datastructures.html

Answer 3

Use del to delete the index of the list you want to remove.

del CNTS[int(np.where(list(np.array_equal(row, c) for row in CNTS))[0])]

CNTS

[array([[[  78, 1202]],

        [[  63, 1202]],

        [[  63, 1187]],

        [[  78, 1187]]]), array([[[ 72, 742]],

        [[ 58, 742]],

        [[ 57, 741]],

        [[ 57, 727]],

        [[ 58, 726]],

        [[ 72, 726]]]), array([[[ 66, 194]],

        [[ 51, 194]],

        [[ 51, 179]],

        [[ 66, 179]]])]